Home Class 11 Applied Maths Chapter 9: Probability Exercise 9.3: Addition Theorem of Probability
Chapter 9 • Unit IV

Exercise 9.3: Addition Theorem of Probability

Comprehensive guide and practice questions for CBSE Class 11 Applied Mathematics curriculum (2026-27).

9.3 Addition Theorem of Probability: General Additivity, Mutual Exclusivity, and De Morgan's Rules

In Applied Mathematics, operations management, and multi-risk underwriting, events frequently overlap. When calculating the probability that a financial portfolio experiences credit default OR liquidity shortfall, simple addition causes double counting. The Addition Theorem of Probability provides the exact algebraic adjustment.

Fundamental Formulae & Theorems

1. Two-Event Addition Theorem

For any two arbitrary events \(A\) and \(B\) associated with a sample space \(S\):

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Mutually Exclusive Case: If \(A \cap B = \emptyset\), then \(P(A \cap B) = 0\), yielding: \[ P(A \cup B) = P(A) + P(B) \]

2. Essential Set-Theoretic Probability Corollaries

  • Probability of "Only \(A\)" (Event \(A\) but not \(B\)): \[ P(A - B) = P(A \cap B') = P(A) - P(A \cap B) \]
  • Probability of "Only \(B\)" (Event \(B\) but not \(A\)): \[ P(B - A) = P(B \cap A') = P(B) - P(A \cap B) \]
  • Probability of "Exactly One" of the Two Events (Symmetric Difference): \[ P(\text{Exactly 1}) = P(A \Delta B) = P(A \cap B') + P(B \cap A') = P(A) + P(B) - 2P(A \cap B) = P(A \cup B) - P(A \cap B) \]
  • Probability of "Neither \(A\) nor \(B\)" (De Morgan's First Law): \[ P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \]
  • Probability that "At Least One Event Does Not Occur" (De Morgan's Second Law): \[ P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B) \]

3. Three-Event Addition Theorem

For any three events \(A, B, C\):

\[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) \]

Comprehensive Applied Example: Student Placement Assessment

Scenario: Two students, Amit and Priya, appear for an investment banking placement interview. The probability that Amit is selected is \(P(A) = 0.5\), the probability that Priya is selected is \(P(B) = 0.6\), and the probability that both are selected is \(P(A \cap B) = 0.3\).

Find the probability that:

  1. At least one of them is selected.
  2. Only Amit is selected.
  3. Exactly one of them is selected.
  4. Neither of them is selected.

Solution:

  1. At least one is selected: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.6 - 0.3 = 0.80. \]
  2. Only Amit is selected (\(A\) but not \(B\)): \[ P(A \cap B') = P(A) - P(A \cap B) = 0.5 - 0.3 = 0.20. \]
  3. Exactly one of them is selected: \[ P(\text{Exactly 1}) = P(A \cup B) - P(A \cap B) = 0.80 - 0.30 = 0.50. \] Verification via components: \(P(A \cap B') + P(B \cap A') = (0.5 - 0.3) + (0.6 - 0.3) = 0.2 + 0.3 = 0.50\). Matches!
  4. Neither is selected: \[ P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.80 = 0.20. \]
Previous: Ex 9.2 Chapter Hub Next: Ex 9.4